It's well-known that over $\mathbb F_q$ every smooth projective conic $C$ is isomorphic to a projective line. So the formula for the motivic zeta-function $Z_{mot}(C)$ is evident since $S^n\mathbb P^1 \simeq \mathbb P^n$.

But what can one say when the finite field in the formulation of this question is replaced by an arbitrary field? Are there any references which can help me to determine the answer?


$S^nC$ will equal $\mathbb P^n$ for $n$ even and a Severi-Brauer variety with the same Brauer class as $C$ for $n$ odd. In the $n$ odd case, its class in the Grothendieck group will be $[C] ( 1+ L^2 + \dots + L^{n-1})$.

Combining the even and odd cases, you get a formula for the motivic zeta function.

For details, and a much more general result, see Zeta Functions of Curves with No Rational Points by Daniel Litt.

  • $\begingroup$ Hi, thank you very much! The desired result sounds quite elementary: isn't there an elementary proof of it? The Litt's proof doesn't seem to me to be so.... $\endgroup$
    – John S.
    Aug 3 at 13:42
  • $\begingroup$ @JohnS. The difficulty in Litt's proof is mainly to do with working with Brauer-Severi varieties over a general base, rather than, as in your case, Brauer-Severi varieties over a point. One approach would be to extract the subset of the proof which is necessary in this case. For example, I think you need Theorem 24 (in the arXiv version, at least), but the proof of that isn't hard, and when you have that, you are almost done. $\endgroup$
    – Will Sawin
    Aug 3 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.